The diagram shows a circle of radius 8 cm and centre O. The tangent to the circle at T meets OA produced at R and OB produced at S. The area of minor sector OAB is 57.5 cm^2, AR = 6 cm and BS = 4 cm.

1. Substitute the sector area and radius into the sector area formula.

For a sector, using radians,
$\text{Area}=\frac{1}{2}r^2\theta$
Given $\text{Area}=57.5$ and $r=8$,
$57.5=\frac{1}{2}(8)^2\theta$

2. Solve for the angle in radians and round to 3 decimal places.

$57.5=32\theta$
$\theta=\frac{57.5}{32}=1.796875$
$\theta=1.797\text{ radians} \quad \text{(to 3 d.p.)}$

1. Find the lengths $OR$ and $OS$.

Since $OA=OB=8$ cm,

$OR=OA+AR=8+6=14$ cm

$OS=OB+BS=8+4=12$ cm

2. Calculate the area of triangle $ORS$ using the included angle from part (a).

Using $\angle ROS=\angle AOB=1.797$ radians,

$\text{Area of }\triangle ORS=\frac12(14)(12)\sin(1.797)$

$=81.86\ldots\text{ cm}^2$

3. Subtract the sector area to find the shaded area.

The shaded area is the part of $\triangle ORS$ outside sector $OAB$.

$\text{Shaded area}=81.86\ldots-57.5$

$=24.36\ldots\text{ cm}^2$

$\approx 24.4\text{ cm}^2$